Proof that, if the norm of any normed space is (Fréchet) differentiable, then the derivative is continuous. I want to prove the following lemma.

Let $X$ be a normed space and its norm $\|.\|$ be Fréchet differentiable on $X \backslash \{0\}$. Then the derivative is continuous there.

I've got this from a book and at first I was very confused because by definition the "derivative" is a linear and continuous operator:

Let $\|.\|$ be Fréchet differentiable in $x\in X \backslash \{0\}$. This means that
$$
\lim_{h\to 0} \frac{\|x+hy\|-\|x\|}{h} = Ty
$$
for some bounded linear operator $T$ on $X \backslash \{0\}$ and that the convergence is uniform for all $y$ with $\|y\|\leq1$.

But I guess what this means is that the map $x\mapsto T$ is supposed to be continuous?
Based on this meaning I tried to approach this head-on with the definition, but apart from it becoming quite tedious, I wasn't able to yield anything meaningful. So either I'm overlooking some trick that makes this simpler or I have to employ some theoretical background that I don't have.
Can someone help me out with this?
 A: The answer is no. If you had a Hilbert space then maybe it would be yes.
Consider the space $\mathrm{X} = \mathscr{K}(\mathbf{R}^\mathbf{N})$ of all real-valued sequences that are eventually zero endowed with the norm $\| (x_m) \| = \sup\limits_{m \in \mathbf{N}} |x_m|.$ This makes of $\mathrm{X}$ a normed space (if you want $\mathrm{X}$ to be a complete normed space then change it to all sequences that converge to zero but I will not prove that such $\mathrm{X}$ is complete, in any case, the theory of derivatives in normed spaces do not need completeness).
Consider the sequence $a_n \in \mathrm{X}$ given by $a_n = (\frac{1}{n} \delta_{n, m})_{m \in \mathbf{N}},$ using the standard Kronecker's delta. Notice that $\| a_n \| = \dfrac{1}{n}.$ If $h = (h_m)_{m \in \mathbf{N}}$ has norm $< \dfrac{1}{4n}$ then
$$
\|a_n + h\| = \dfrac{1}{n} + h_n = \|a_n\| + h_n.
$$
Therefore, the derivative of $u( \cdot ) = \| \cdot \|$ at $a_n$ is nothing else that the continuous linear function $u'(a_n):\mathrm{X} \to \mathbf{R}$ given by $u'(a_n) \cdot (x_m)_{m \in \mathbf{N}} = x_n,$ which is the projection onto the $n$th axis of $\mathbf{R}^\mathbf{N}.$ Clearly, $\| u'(a_n) \| = 1.$ Therefore, $a \mapsto u'(a)$ from $\mathrm{X}$ into $\mathrm{Lin}(\mathrm{X}, \mathbf{R})$ is not continuous. Q.E.D.
